An arc surgery enlarges a link Seifert matrix by two rows and columns

The two surgery disks and a narrow strip in $F$ joining them have a disk neighbourhood $D$. Replacing the disks by the lateral annulus of the cylinder changes this neighbourhood into a once-punctured torus $P$ with the same boundary $\mathrm{\partial }D$. Write $F_0=\overline{F\setminus D}$. Any old homology class can be represented in $F_0$: replace each of its finitely many arcs through $D$ by an arc along $\mathrm{\partial }D$, which changes the cycle by a boundary in the disk. Take open collar neighbourhoods in $F$ and $F^{\mathrm{\prime }}$ so that $F_0$ and respectively $D$ or $P$ form open covers whose intersection retracts onto $\mathrm{\partial }D$. In the Mayer–Vietoris sequence [F3], the map from $H_0(\mathrm{\partial }D;\mathbb{Z})=\mathbb{Z}$ to the direct sum of the two $H_0$ groups is $n\mapsto (n,-n)$ and is injective. For $F=F_0\cup D$, the $H_1$ map from the intersection circle has image generated by $([\mathrm{\partial }D],0)$, since $H_1(D;\mathbb{Z})=0$. For $F^{\mathrm{\prime }}=F_0\cup P$, the same image is generated by $([\mathrm{\partial }D],0)$ because the oriented boundary of $P$ is zero in $H_1(P;\mathbb{Z})$, while $H_1(P;\mathbb{Z})\cong {\mathbb{Z}}^2$ on its longitudinal and meridional curves. Thus $H_1(F^{\mathrm{\prime }})\cong H_1(F)\oplus {\mathbb{Z}}^2$, with the old classes as a direct summand. Choose the added basis classes $b,c$ from those two curves; together with the old basis they are an integral basis.

The surgery retains the old oriented surface outside the two attaching disks, so its positive normal there is unchanged and the old matrix block remains $V$. Use product coordinates $(t,w)$ on the cylinder, with $w\in D^2$, and take $c=\{t_0\}\times \mathrm{\partial }{D}^2$. Let $b$ follow one longitudinal side of the tube and return through $F_0$; it crosses each cross-section disk once. The disks $\{t_0\}\times D^2$ and their sufficiently close parallel disks miss every old basis curve and its push-off, so [F4] gives zero for the two Seifert pairings between $c$ and every old basis curve. Along the lateral annulus, the positive normal points either out of or into the solid cylinder. If it points out, the radius-one cross-section disk misses $b^{+}$, giving $V(b,c)=0$, while the disk of slightly larger radius bounds $c^{+}$ and meets $b$ once; it gives $\mathrm{lk}(b,c^{+})=+1$ after orienting $c$, hence $V(c,b)=+1$ by the symmetry in [F4]. If it points in, the radius-one disk meets $b^{+}$ once, giving $V(b,c)=+1$ after orienting $c$, while the disk of slightly smaller radius bounds $c^{+}$ and misses $b$, giving $\mathrm{lk}(b,c^{+})=0$ and therefore $V(c,b)=0$. These intersection statements follow from the explicit single transverse point in the product coordinates and [F4]. In either case the concentric parallel meridians have $V(c,c)=0$. The remaining entries involving $b$ are integers $x$, $y$, and $z=\mathrm{lk}(b^{+},b)$. Thus in the ordered basis $(a_1,\dots ,a_m,b,c)$ the matrix has exactly one of the two displayed forms.